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Zinc-EDTA Titration
You would like to perform a titration of 50.00 mL of a 1.00 x 10-4 M Zn2+ solution with a 1.00 x 10-4 M EDTA solution.
What is pZn at the equivalence point?
Log Kf for the ZnY2- complex is 16.5.
Both solutions are buffered to a pH of 10.0 using a 0.100M ammonia buffer.
R. Corn Chem M3LC Spring 2013
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We use an ammonia buffer solutionfor a reason...
αY4- = 0.355 at pH =10.
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Solubility product Ksp = 3.0×10−17
Zn2+ + 2OH- → Zn(OH)2(s)
Ksp = [Zn2+][OH -]2
At pH=10, [Zn2+] = ??
Zinc Hydroxide
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Solubility product Ksp = 3.0×10−17
Zn2+ + 2OH- → Zn(OH)2(s)
Ksp = [Zn2+][OH -]2
At pH=10, [Zn2+] = Ksp /[OH -]2
Zinc Hydroxide
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Solubility product Ksp = 3.0×10−17
Zn2+ + 2OH- → Zn(OH)2(s)
Ksp = [Zn2+][OH -]2
At pH=10, [Zn2+] = Ksp /[OH -]2
Zinc Hydroxide
= (3.0×10−17) /(1.0×10−4)2
= 3.0×10−9 M
This is the maximum free Zinc concentration.
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Ammonia Buffer Solution
At pH=10, [NH3] = ??
[NH3]total = 0.100M
NH3 + H2O <=> NH4+ + OH - pKb = 4.74
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Ammonia Buffer Solution
At pH=10:
[NH3]total = 0.100M
NH3 + H2O <=> NH4+ + OH -
[NH3] = 0.0846 M
pKb = 4.74
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Zinc - Ammonia Complexation
At [NH3] = 0.0846M, αZn2+ = 1.61 x 10-5
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Zinc - Ammonia Complexation
At a TOTAL Zinc concentration of 1.00 x 10-4 M:
[Zn2+] = αZn2+ [Zn2+]tot
[Zn2+] = (1.61 x 10-5)(1.00 x 10-4)
[Zn2+] = 1.61 x 10-9 M
Therefore: no precipitation!
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EDTA Titration
You would like to perform a titration of 50.00 mL of a 1.00 x 10-4 M Zn2+ solution with a 1.00 x 10-4 M EDTA solution.
What is pZn at the equivalence point?
Log Kf for the ZnY2- complex is 16.5. Both solutions are buffered to a pH of 10.0 using a 0.100M ammonia buffer.
The alpha fraction for Y4- is 0.355 at a pH of 10.0.The alpha fraction for Zn2+ is 1.61 x 10-5.
Once more, with feeling:
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EDTA Titration
You would like to perform a titration of 50.00 mL of a 1.00 x 10-4 M Zn2+ solution with a 1.00 x 10-4 M EDTA solution.
Total moles of Zinc = ??
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EDTA Titration
You would like to perform a titration of 50.00 mL of a 1.00 x 10-4 M Zn2+ solution with a 1.00 x 10-4 M EDTA solution.
Total moles of Zinc = (1.00 x 10-4 M)(0.050L)
= 5.0 x 10-6 moles
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EDTA Titration
You would like to perform a titration of 50.00 mL of a 1.00 x 10-4 M Zn2+ solution with a 1.00 x 10-4 M EDTA solution.
Total moles of Zinc = 5.0 x 10-6 moles
Equivalence point volume = ??
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EDTA Titration
You would like to perform a titration of 50.00 mL of a 1.00 x 10-4 M Zn2+ solution with a 1.00 x 10-4 M EDTA solution.
Total moles of Zinc = 5.0 x 10-6 moles
Equivalence point volume = 0.100L
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EDTA Titration
You would like to perform a titration of 50.00 mL of a 1.00 x 10-4 M Zn2+ solution with a 1.00 x 10-4 M EDTA solution.
Total moles of Zinc = 5.0 x 10-6 moles
Equivalence point volume = 0.100L
[ ZnY2- ] = ??
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EDTA Titration
You would like to perform a titration of 50.00 mL of a 1.00 x 10-4 M Zn2+ solution with a 1.00 x 10-4 M EDTA solution.
Total moles of Zinc = 5.0 x 10-6 moles
Equivalence point volume = 0.100L
[ ZnY2- ] = 5.0 x 10-5 M
We assume a stoichiometric reaction.
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EDTA Titration
You would like to perform a titration of 50.00 mL of a 1.00 x 10-4 M Zn2+ solution with a 1.00 x 10-4 M EDTA solution.
Total moles of Zinc = 5.0 x 10-6 moles
Equivalence point volume = 0.100L
[ ZnY2- ] = 5.0 x 10-5 M
[ Zn2+ ] = ??
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[ ZnY2- ] = 5.0 x 10-5 M
We assumed a stoichiometric reaction. But actually, there is a little bit of free (uncomplexed) EDTA and free (uncomplexed) Zinc in solution.
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[ ZnY2- ] = 5.0 x 10-5 M
We assumed a stoichiometric reaction. But actually, there is a little bit of free (uncomplexed) EDTA and free (uncomplexed) Zinc in solution.
αZn2+ = 1.61 x 10-5
Log Kf = 16.5. αY4- = 0.355
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[ ZnY2- ] = 5.0 x 10-5 M
We assumed a stoichiometric reaction. But actually, there is a little bit of free (uncomplexed) EDTA and free (uncomplexed) Zinc in solution.
1.66 x 10-8 M
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EDTA Titration
You would like to perform a titration of 50.00 mL of a 1.00 x 10-4 M Zn2+ solution with a 1.00 x 10-4 M EDTA solution.
Total moles of zinc = 5.0 x 10-6 moles
Equivalence point volume = 0.100L
[ ZnY2- ] = 5.0 x 10-5 M 1.66 x 10-8 M
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EDTA Titration
You would like to perform a titration of 50.00 mL of a 1.00 x 10-4 M Zn2+ solution with a 1.00 x 10-4 M EDTA solution.
Total moles of zinc = 5.0 x 10-6 moles
Equivalence point volume = 0.100L
[ ZnY2- ] = 5.0 x 10-5 M
= (1.61 x 10-5) (1.66 x 10-8 M)
[ Zn2+ ] = 2.59 x 10-13 M pZn = 12.6
We are done!
2.35 x 10-6 M